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3 years ago

At what location does gravity play a role in moving tectonic plates

Physics

2 answers:

Ksju [112]3 years ago

8 0

At the edge of the plates.

Lena [83]3 years ago

7 0

Answer:

In subduction of one plate over another on the basis of weight

Explanation:

When two plates moving and colliding, the plate with higher mass will subduct below the plate with lower weigh and lighter would remain above. generally continental types of plates are lighter in weight and oceanic plates are heavier.

Ex:- The collision of Eurasian plate and Indian plate. Indian plate is below and Eurasian plate above and formation of Himalaya

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Two stationary point charges of 3.00 nC and 2.00 nC are separated by a distance of 50.0 cm. An electron is released from rest at

andrew-mc [135]

Answer:

1. the electric potential energy of the electron when it is at the midpoint is - 2.9 x $10^{-17}$ J

2. the electric potential energy of the electron when it is 10.0 cm from the 3.00 nC charge is - 5.04 x $10^{-17}$ J

Explanation:

given information:

$q_{1}$ = 3 nC = 3 x $10^{-9}$ C

$q_{2}$ = 2 nC = 2 x $10^{-9}$ C

r = 50 cm = 0.5 m

the electric potential energy of the electron when it is at the midpoint

potential energy of the charge, F

F = k $\frac{q_{e}q}{r}$

where

k = constant (8.99 x $10^{9} Nm^{2} /C^{2}$ )

electron charge, $q_{e}$ = - 1.6 x $10^{-19}$ C

since it is measured at the midpoint,

r = $\frac{0.5}{2}$

= 0.25 m

thus,

F = $F_{1}+ F_{2}$

= k $\frac{q_{e} q_{1} }{r}$ + k $\frac{q_{e} q_{2} }{r}$

= $\frac{kq_{e} }{r}$ ( $q_{1} +q_{2}$ )

= (8.99 x $10^{9}$ )( - 1.6 x $10^{-19}$ )(3 x $10^{-9}$ +2 x $10^{-9}$ )/0.25

= - 2.9 x $10^{-17}$ J

the electric potential energy of the electron when it is 10.0 cm from the 3.00 nC charge

$r_{1}$ = 10 cm = 0.1 m

$r_{2}$ = 0.5 - 0.1 = 0.4 m

F = k $\frac{q_{e} q_{1} }{r}$ + k $\frac{q_{e} q_{2} }{r}$

= $kq_{e}$ ( $\frac{q_{1} }{r_{1} }$ + $\frac{q_{2} }{r_{2} }$ )

= (8.99 x $10^{9}$ )( - 1.6 x $10^{-19}$ )(3 x $10^{-9}$ /0.1+2 x $10^{-9}$ /0.4)

= - 5.04 x $10^{-17}$ J

3 0

3 years ago

A cylindrical metal specimen having an original diameter of 11.77 mm and gauge length of 46.1 mm is pulled in tension until frac

Marta_Voda [28]

Answer:

% reduction in area = 54.26 %

percentage elongation = 43.16 %

Explanation:

a) percentage reduction in area $= \frac{(A_1 - A_2)}{A_1 } * 100$

$A_1 =\pi r^2 = \pi * 5.88^2=108.8 mm2$

$A_2=\pi * 3.98^2= 49.97 mm2$

% reduction in area = $\frac{(108.8 - 49.97)}{108.8} * 100 = 54.26 %$

b)percentage elongation $= \frac{(66 - 46.1)}{46.1} *100 = 43.16 %$

5 0

3 years ago

A person is pulling their 20 kg luggage using the luggage handle. The handle is at an angle of 56 degrees above the horizontal.

rusak2 [61]

Answer:

The answer to your question is: a = 1.99 m/s²

Explanation:

Data

mass = 20 kg

angle = 56°

Force = 71 N

horizontal acceleration = ?

Process

Find the horizontal force

cos Ф = adjacent side / hypotenuse

adjacent side = hypotenuse x cosФ

adjacent side = 71 x cos 56

a.s. = 39.70 N

Newton's second law

F = ma

a = F/m

a = 39.7 / 20

a = 1.99 m/s²

4 0

3 years ago

3) All numbers that are divisible by both 3 and 5 are also divisible by 10. Which of the following numbers can be used to show t

zlopas [31]

Answer:

answer E

Explanation:

cuz its 45 is divisible by 3 and 5 but not 10

8 0

2 years ago

What does Newton's law all about?<br>

KengaRu [80]

Newton's law is all about motion

5 0

2 years ago